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(C) Given that the concentration of donor atoms (Arsenic) is $N_D = 4.5 	imes 10^{21} \ m^{-3}$.
Since Arsenic is a pentavalent impurity, the number

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$9 	imes 10^{10}$

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(C) Given that the concentration of donor atoms (Arsenic) is $N_D = 4.5 	imes 10^{21} \ m^{-3}$.
Since Arsenic is a pentavalent impurity, the number of electrons $n_e \approx N_D = 4.5 	imes 10^{21} \ m^{-3}$.
The intrinsic carrier concentration is $n_i = 1.5 	imes 10^{16} \ m^{-3}$.
Using the law of mass action, $n_e \cdot n_h = n_i^2$, the number of holes $n_h$ is given by:
$n_h = \frac{n_i^2}{n_e} = \frac{(1.5 	imes 10^{16})^2}{4.5 	imes 10^{21}} = \frac{2.25 	imes 10^{32}}{4.5 	imes 10^{21}} = 0.5 	imes 10^{11} = 5 	imes 10^{10} \ m^{-3}$.
The ratio of the number of electrons to the number of holes is:
$\frac{n_e}{n_h} = \frac{4.5 	imes 10^{21}}{5 	imes 10^{10}} = 0.9 	imes 10^{11} = 9 	imes 10^{10}$.

The number of silicon atoms per $m^3$ is $5 \times 10^{28}$. This is doped with $4.5 \times 10^{21}$ atoms $/ m^3$ of Arsenic. The ratio of the number of electrons to the number of holes after doping is (Take $n_i = 1.5 \times 10^{16} / m^3$)

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